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Is there any way to reduce voltage without using transformer? I have a supply of 24 volts AC @ 1500mA (1.5A) and I want to convert it into 6 volts @ 500mA. Because I have to supply the 6 volts to the 7" compact LCD screen via rectifier. Since, there is no enough space available in the LCD housing to use transformer and I think transformer may also cause screen to damage due to its magnetic fields.

Schematic will be highly appreciated.

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LCD screens generally aren't affected by magnetic fields. You don't have to worry about putting a magnet that close to it. – MBraedley Nov 15 '11 at 18:31
Which solution seems to work? (I suggested several.) – Russell McMahon Nov 15 '11 at 21:42
I mean Resistor dropper + regulator Cheap and nasty solution. – Farid-ur-Rahman Nov 16 '11 at 17:01
LCDs are not affected by magnetic fields, like CRTs were. So I wouldn't worry about it! – Dark Goob Dec 22 '11 at 7:19
up vote 7 down vote accepted

External transformer will work well. If you can find a mains to suitable multiple low voltage output windings transformer you can make your own step down. eg a 12/12/6 can be arranged as a 24:6 which is about right.

SMPS is a good idea due to efficiency.
Slightly more dangerous load safety wise - see below.

eg ye olde MC34063

This would cost a few dollars.
Note that with ANY non isolated circuit protection should be provided that will blow a fuse if the converter fails shirt circuit. Running yuour LCD on 24V would be a bad idea. See fig 24 in the above datasheet that shows an isolated output converter.

enter image description here

Resistor dropper + regulator Cheap and nasty solution that works OK

Note that this is quick and easy and nasty. Not recommended but doable.

Rectify 24VAC to about 34 Volts (!)
Pass this through a 47 ohm 20 Watt resistor (can be several smaller resistors in suitable series / parallel arrangement) and then
connect this to a 5V, 1A regulator such as LM340, 7805 etc.

Resistor max drop = 0.5A x 47R =~ 24 V.
V into regulator = 34-24 = 10V max.

Power in resistor = I^2 x R = 0.25 x 48 = 12 Watt.
Use much larger rating eg 20W+

You can tailor R so that V in to regulator is about 8V. Less may cause troubles.
You need a cap at regulator input.

Heatsink regulator appropriately.

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Seems to work. Thanks for the support. – Farid-ur-Rahman Nov 15 '11 at 18:29
Would someone seriously waste 12W in a resistor, do supply 2,5W to a circuit? I guess the combination heatsink + beefy resistor might even be more expensive than a SMPS. – 0x6d64 Nov 16 '11 at 9:54
@0x6d64 - Some may use a resistivedropper. I'd avoid it if at all possible BUT resistors are cheap enough and even compact enough. Size it right electrically and heatsink is minimal. Possibly safer than many smps as resistors will fail O/C whereas a smps can well enough fail s/c. Not nice energy wise but very doable. – Russell McMahon Nov 16 '11 at 11:29
Shouldn't this design be isolated? If no, why not? – abdullah kahraman Dec 22 '11 at 8:21
@abdullahkahraman - Please read my answer. Isolation and merits are discussed. As part only of this search for "see fig 24" in the above answer. – Russell McMahon Dec 22 '11 at 11:04

Since you need a 6V DC I would suggest to use a DC/DC converter. Rectify and smooth the 24V, and use a "buck" converter, aka "switcher" or "SMPS" (Switch-Mode Power Supply).
These require only a few parts; National has a Webench web application which helps you with the design, up to the BOM. For instance, Webench shows a design around an LM2674-ADJ with a BOM cost less than USD 2.00 (sic) and an efficiency of 85%.

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Shouldn't this design be isolated? If no, why not? – abdullah kahraman Dec 22 '11 at 8:20
@abdullah - Probably doesn't need isolation, I'm assuming the 24V AC is already separated from the mains by means of a transformer. – stevenvh Dec 22 '11 at 16:57

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